How does load resistance affect RC low pass filter?

I am building a low pass filter. This is the equation I'm using to calculate what frequencies I want to pass though, but I have a couple questions on details.

$$f_c = \frac{1}{2\pi RC}$$

What goes into R? Does the resistance of the load (R2) affect R? Say R2 is fixed, are there any rules for selectring R1, like load matching?

simulate this circuit – Schematic created using CircuitLab

• Use Req=R1//R2 = R and attenuation is R2/(R1+R2) at DC Jul 21 '20 at 6:53
• Jul 21 '20 at 7:03

You can use the Thevenin equivalent circuit to see what your capacitor "sees".

Your capacitor will see an input voltage of Vin*R2/(R1+R2) in series with a resistor equal to (R1||R2). Hence your load will affect your voltage amplitude and also your cut-off frequency. If it is a problem, you may use an op-amp configured as a voltage follower between the output of your filter and your load.

This simple circuit can be solved by inspection instantaneously either by using Thévenin as correctly highlighted by Wheatley but also by using the fast analytical circuits techniques or FACTs. Simply consider the circuit for $$\s=0\$$ and determine the dc transfer function. Then turn the stimulus off - short the input source - and "look" through the capacitor terminals to determine the resistance $$\R\$$ you see in this mode. Multiply the result by the capacitor to get the time constant. For a 1st-order circuit, the inverse of the time constant is the pole $$\\omega_p\$$ you want. The below drawing illustrates the work:

Then assemble the pieces and write the transfer function in a low-entropy form as follows:

If you want to check if the equation is applicable (and what is "R" ?), you should use the definition for the cut-off frequency wo. For this purpose you need the transfer function.

It is a first-order lowpass - hence, the general expression applies:

H(jw)=Ao/(1+jw/wo).

Ao: Transfer function for w=0

wo=1/T with T=time constant of the RC product relevant for discharging the capacitor.

• It would be interesting for me to learn if something is wrong in my answer...has the "downvoter" the courage to reveal his secret? It was my aim not to present the answer but enable the questioner to find it by his own...
– LvW
Jul 21 '20 at 10:35